- #1
Ryoukomaru
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This is a problem I had in a test and almost everyone got different answers for it, we discussed and well, I spotted mistakes in their solutions so I think mine is right but I wanted to check here and also ask if there is an easier/faster way to do it.
There is a container that is similar to the bottom part of a cone which is cut into half. Top radius is [tex]10cm[/tex] and bottom is [tex]20cm[/tex]. And the volume of this container is [tex]500cm^{3}[/tex]. What is the length of the slanted side ?
So what I did was, first I drew this. ;P (See attachment)
Then what I did was to write an equation for [tex]V_2[/tex] in terms of [tex]V_1[/tex] and [tex]V_T[/tex]
By solving the equation for height, I got [tex]h=1.5915[/tex]
Then I used Pythagoras' theorem to find the length of the slanted side and it comes to:
[tex]10^2+2h^2=s^2=>100+3.183^2=s^2=> s=10.4944[/tex]
Is this correct ? And is there a formula to find the volume of this shape ?
There is a container that is similar to the bottom part of a cone which is cut into half. Top radius is [tex]10cm[/tex] and bottom is [tex]20cm[/tex]. And the volume of this container is [tex]500cm^{3}[/tex]. What is the length of the slanted side ?
So what I did was, first I drew this. ;P (See attachment)
Then what I did was to write an equation for [tex]V_2[/tex] in terms of [tex]V_1[/tex] and [tex]V_T[/tex]
By solving the equation for height, I got [tex]h=1.5915[/tex]
Then I used Pythagoras' theorem to find the length of the slanted side and it comes to:
[tex]10^2+2h^2=s^2=>100+3.183^2=s^2=> s=10.4944[/tex]
Is this correct ? And is there a formula to find the volume of this shape ?